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For a function $f(x)$ on $\mathbb{R}$, its Fourier Transform is given by

$$F(\xi) = \frac{1}{2\pi}\int_{R} f(x)e^{-i \xi x}dx $$.

Then, what is the Fourier Transform of the wave function $f(x) = e^{i \xi x} $ itself? If I simply plug it in, I would get $\infty$. What does this mean?

Also, if $f(x) = e^{i \eta x}$, where $\eta \not = \xi$, what is its Fourier Transform?

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    $\begingroup$ $\delta(\xi-\eta)$ $\endgroup$ – Jacky Chong May 12 '17 at 2:23
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    $\begingroup$ First, $f(x)$ should not depend on the transform variable $\xi$. If $f(x)=e^{I\eta x}$ and $\eta \in \mathbb{R}$, then the Fourier transform of $f$ is $\delta(\eta -\xi)$ in terms of the Dirac Delta. $\endgroup$ – Mark Viola May 12 '17 at 2:23
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first your definition of fourier transform is wrong.It should be : $$F(\zeta)=\int f(x)e^{-i\zeta x}dx $$ where f(x) is function of x only .By using fourier transform we are converting from x domain to $\zeta$ domain. now for the second case the fourier transform will be : $2\pi\delta(\zeta-\eta)$

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